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Elina [12.6K]
3 years ago
5

How do I solve X to the second power -55 equals 26

Mathematics
2 answers:
guapka [62]3 years ago
8 0

Answer:

Solutions x= 9 and -9

Step-by-step explanation:

x^2-55=26

x^2=55+26

x^2=81

(take square root of both sides)

x=9

(squaring negatives is the same as squaring the same positive)

x also=-9

sammy [17]3 years ago
3 0

Answer:

x= 9 x= -9

Step-by-step explanation:

Equation: x^2-55=26

Move -55 to the other side - x^2=55+26 = X^2 = 81

Solve - x^2=81 - 9^2 = 81 and -9^2=81 (a negative squared becomes a positive)

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find the area of field in a form of trapezium whose parallel sides are 10 and 12 and perpendicular distance between them is 6 me
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Answer:

66 m²

Step-by-step explanation:

A = ½ (a + b) h

A = ½ (10 + 12) (6)

A = 66

The area is 66 m².

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Explain how the figure illustrates that 6(9) = 6(5) + 6(4)
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(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

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